Einstein brother-in-law elevator

  • #1
member 728827
TL;DR Summary
As personal curiosity, I want to calculate which is the difference in "travelled height" between a photon that goes across the width of an elevator - which is more or less 2[m] in my country - and a tiny mass particle that free-falls starting at the same "height" as the photon origin, and is released exactly at the same time that the photon is shot from the other side.
As personal curiosity, I want to calculate which is the difference in "travelled height" between a photon that goes across the width of an elevator - which is more or less 2[m] in my country - and a tiny mass particle that free-falls starting at the same "height" as the photon origin, and is released exactly at the same time that the photon is shot from the other side.

I'll make some assumptions in order to make this race "fair enough". Hope I don't make many mistakes and correct me if I'm wrong. We'll see...

First of all, thanks to the sponsors of this "tough" experiment: Coke© light (supplying the laser assembly) and Nanomass© dynamics for the test particle (which they sell by the weight and are dirty cheap).

Also, due the lack of budget, the test will be conducted in Earth, simplifying and considering the Schwarzschild metric solution to the Einstein Field Equations, so Earth and surrounding spacetime are not flat.

This defines the "Schwarzschild radius", and will be used in the metric:
$$[1]\quad R_s=\frac{2GM}{c^2}$$

The Schwarzschild metric itself:
$$[2]\quad ds^2=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\theta^2+r^2sin^2(\theta)d\phi^2$$

Both the photon and the test particle will be in "free-fall" once released, and thus will follow a Geodesic trajectory. To obtain the motion equations, first I'll begin with the Euler-Lagrange condition, because such trajectories will maximize some Lagrangian function.

$$[3]\quad (x^0,x^1,x^2,x^3)=(ct,r,\theta,\phi) \qquad \dot{x}^\mu=\frac{\partial x^\mu}{\partial λ} \qquad \frac{\partial}{\partial λ}(\frac{\partial\mathscr{L}}{\partial\dot{x}^\mu})-\frac{\partial\mathscr{L}}{\partial x^\mu}=0$$

Defining the Lagrangian as the derivative of the line interval by a suitable parameter λ:

$$[4]\quad \mathscr{L}=(\frac{ds}{dλ})^2=-c^2(1-\frac{R_s}{r})\dot{t}^2+\frac{\dot{r}^2}{(1-\frac{R_s}{r})}+r^2\dot{\theta}^2+r^2sin^2(\theta)\dot{\phi}^2$$

Applying the Euler-Lagrange rule to the coordinates, we get the following general conditions that the Geodesic trajectories must follow:

- We can consider the trajectory contained in a plane. Let's choose for sake of simplicity $\theta=\pi/2$:
$$[5]\quad \mathscr{L}(\theta) \Rightarrow 2r^2\ddot{\theta}-2r^2\dot{\theta}^2sin(\theta)cos(\theta)=0 \Rightarrow
\theta=\pi/2$$

- There're a two conserved quantities (constants of motion) to be determined for a Geodesic path:

$$[6]\quad \mathscr{L}(ct) \Rightarrow -2c^2(1-\frac{R_s}{r})\ddot{t}=0 \Rightarrow (1-\frac{R_s}{r})\dot{t}=k$$
$$[7]\quad \mathscr{L}(\phi)\Rightarrow 2r^2sin^2(\theta)\ddot{\phi}=0\Rightarrow r^2sin^2(\theta)\dot{\phi}=l$$

Dividing [7] into [6], we get rid of the derivatives with respect to parameter λ, and can write the constant of motion directly as a consequence of the coordinate components:

$$[9]\quad b=\frac{l}{k}=\frac{r^2\dot{\phi}}{(1-\frac{R_s}{r})\dot{t}}=\frac{r^2}{(1-\frac{R_s}{r})}\frac{d\phi}{dt}$$

Combining all this stuff, we obtain our first general equation of motion for a Geodesic path:

$$[10]\quad \Rightarrow \frac{d\phi}{dt}=\frac{1}{r^2}(1-\frac{R_s}{r})\cdot b$$

Now, we have to distinguish between a test particle (with mass), and a (massless) photon.

For light, and using the original metric equation [2]:

$$[11]\quad photon: ds^2=0=-c^2(1-\frac{R_s}{r})dt^2+\frac{dr^2}{(1-\frac{R_s}{r})}+r^2d\phi^2$$

And then, here we have our second equation of motion, related to a constant b to be determined afterwards:

$$[12]\quad (\frac{dr}{dt})^2=-\frac{b^2}{r^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

For the constant of motion, if consider the launch angle ##\varphi## - from the horizontal to the initial direction -, and the primed coordinates (r' and t') for a local observer which always measures (as Einstein likes) the speed of light as c, we have:

$$[13]\quad t=t_0=0,r=r_0 \qquad \frac{dr'}{dt'}|_{t0}=c \cdot sin(\varphi)$$
$$c^2dt'^2=c^2(1-\frac{R_s}{r})dt^2 \qquad dr'^2=\frac{dr^2}{(1-\frac{R_s}{r})}\Rightarrow \frac{dr'}{dt'}=\frac{1}{(1-\frac{R_s}{r})} \cdot \frac{dr}{dt}$$
$$\frac{dr}{dt}|_{t_0}=(1-\frac{R_s}{r_0}) \cdot c \cdot sin(\varphi)$$

$$[14]\quad (1-\frac{R_s}{r_0})^2\cdot c^2 \cdot sin^2(\varphi)=-\frac{b^2}{r_0^2}(1-\frac{R_s}{r_0})^3+c^2(1-\frac{R_s}{r_0})^2$$
$$\Rightarrow b^2=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot \left\{ 1-sin^2(\varphi) \right\}=\frac{c^2r_0^2}{(1-\frac{R_s}{r_0})} \cdot cos^2(\varphi)$$

Now for the test particle. We do the same, except that is not a null but a timelike Geodesic:

$$[15]\quad \mathscr{L}=(\frac{ds}{d\tau})^2=-c^2=-\frac{c^2k^2}{(1-\frac{R_s}{r})}+\frac{(\frac{dr}{d\tau})^2}{(1-\frac{R_s}{r})} \qquad \theta=\pi/2 \quad d\theta=0 \quad d\phi=0$$

$$[16]\quad -c^2(1-\frac{R_s}{r})=-c^2k^2+(\frac{dr}{d\tau})^2$$

$$[17]\quad \frac{dr}{d\tau}=\frac{dr}{dt}\frac{dt}{d\tau}=\frac{dr}{dt}\frac{k}{(1-\frac{R_s}{r})}$$

The second equation of motion for a mass particle:
$$[18]\quad (\frac{dr}{dt})^2=-\frac{c^2}{k^2}(1-\frac{R_s}{r})^3+c^2(1-\frac{R_s}{r})^2$$

From the initial condition, we can get the constant of motion:
$$[19]\quad t=t_0=0,r=r_0 \qquad \frac{dr}{dt}|_{t_0}=0 \qquad \Rightarrow k=±\sqrt{1-\frac{R_s}{r_0}}$$

As for the setup of the test, if we shoot the photon with an initial non-zero launch angle ##\varphi## (inwards), then it will have obviously an initial inward geometric radial velocity, and being c a big number seems unfair to compare then the radial travelled distance of the photon with a poor test particle that starts quiet and still.

So, we'll shot the photon tangentially (##\varphi=0##). This means that if the path of light were unaffected by Gravity, would hit the other side of the elevator at a higher position (i.e., if the starting radius is r_0, would reach the other side at a radius r_1>r_0). In a nutshell, we don't point the laser towards the test particle, but a little above it.

For the test particle, it's starting position will be r_0, and so both the photon and the test particle begin at the same distance (radius) from the center of the planet.

That's fair enough for me.

With a simple Python code (attached), I obtain the following:

FINAL RESULT:
r1-rend=6.556e-16[m] r0-rend_m=2.185e-16[m] tend=6.671e-09

where r1 is the point-of-impact radius of a straight line with a length of 2[m] (we fix as condition the arc covered to be L/r_0), rend if the final radius of the photon position, rend_m is the final radius of the test particle position, and tend is the total flight time (in the coordinate time). So the radial distance displacement for the photon is 3.0004.. times greater than the radial distance displacement of the test particle.

Python code:
from mpmath import *
import numpy as np
import matplotlib.pyplot as plt
import math
from datetime import datetime

mp.dps = 100
mp.pretty = True

c=299792458        # speed of light
c2=pow(c,2)        # square of speed of light
G=mpf(6.6743e-11)  # Gravitional constant
M=mpf(5.9722e+24)  # Earth's mass
r0=mpf(6371000)    # Earth's radius
L=mpf(2)           # Elevator's width

alp=L/r0

Rs=(2*G*M)/c2
Rs2=pow(Rs,2)
r0s=1-Rs/r0

b2=((pow(r0,2)*c2)/r0s)
b=sqrt(b2)

def zeq(r):
    return c2-(b2/pow(r,2))*(1-Rs/r)

def fph_dt_dr(r):
    zr=zeq(r)

    if (zr>0):
        return 1/((1-Rs/r)*sqrt(zr))
    else:
        raise Exception('NEGATIVE SQRT! r=%5.3e[m]'%(r))

def fph_dphi_dt(r):
    return (b/pow(r,2))*(1-Rs/r)

def fph_dphi_dr(r):
    zr=zeq(r)

    if (zr>0):
        return b/sqrt(r*(c2*pow(r,3)-b2*(r-Rs)))
    else:
        raise Exception('NEGATIVE SQRT! r=%5.3e[m]'%(r))

def fmp_dt_dr(r):
    return -1/(c*(1-Rs/r)*sqrt(1-(1-Rs/r)/r0s))


print('')

r1=r0/cos(alp)

[rend] = findroot([lambda r: quad(fph_dphi_dr,[r0,r])-alp],(r1))
tend = quad(fph_dt_dr,[r0,rend])

d=pow(tend,2)*G*M/(2*pow(r0,2))  # initial guess
[rend_m] = findroot([lambda r: quad(fmp_dt_dr,[r0-1e-51,r])-tend],(r0-d))

print('')   
print('FINAL RESULT:')
print('r1-rend=%5.3e[m] r0-rend_m=%5.3e[m] tend=%9.3e[s]'%(r1-rend,r0-rend_m,tend))
 
  • Like
Likes Dale
Physics news on Phys.org
  • #2
Note that the equations of motion for this case can be found in textbooks. Carroll's lecture notes equations 7.43, 7.44, 7.47 and 7.48 are there, and you can check your results.

There is a closed-form expression for ##t(r)## for radial motion. Wolfram Alpha should be able to determine it given ##dr/dt## in that case. Sadly you cannot get ##r(t)##, but it may provide a better way of calculating the fall distance of the mass.

You can sanity check your results with Newtonian gravity. Light travels ##l=2\mathrm{m}## in ##\delta t=l/c## seconds. It will rise by approximately ##l^2/2R_E## compared to its initial altitude. The mass will fall approximately ##\frac 12g\delta t^2## in that time. You won't be far wrong with this - how much does it disagree with your detailed calculations?
 
  • Like
Likes member 728827
  • #3
Lluis Olle said:
the difference in "travelled height" between a photon that goes across the width of an elevator - which is more or less 2[m] in my country - and a tiny mass particle that free-falls starting at the same "height" as the photon origin, and is released exactly at the same time that the photon is shot from the other side.
If I'm understanding you correctly, you want to release a test particle and a light pulse (the word "photon" is not appropriate in a classical context, you're really firing a very short pulse of light) at the same height in an "elevator", with the test particle starting at rest relative to the elevator and the light pulse released pointing horizontally, and watch how far they both "fall" in the time it takes the light to travel 2 meters across the elevator.

Assuming that my understanding is correct, you are doing an awful lot of work to derive a result that the equivalence principle can tell you immediately: the fall distances will be the same, i.e., the difference in "travelled height" will be zero. Just do the analysis in a local inertial frame centered on the elevator. In that frame, the light pulse and the test object are both in free fall, and the test particle will simply remain at rest in the local inertial frame while the light pulse will travel horizontally across it, with both staying at the same "vertical" coordinate. So in the local inertial frame the "fall distance" is zero for both test particle and light pulse. The elevator itself is accelerating upward relative to the local inertial frame, so the "fall distance" relative to the elevator is just the distance the elevator moves upward in the 6.6 nanonseconds it takes for the light pulse to travel 2 meters.
 
  • Like
Likes ersmith
  • #4
I attach an extremely exaggerated figure, to better understand the scenario.

The laser is fixed at point A, at the left side of the elevator. The test particle is located at point B. Both laser origin and test particle are at the same radial coordinate r0.

For reasons given in the thread, the laser is pointed towards C.

At t0, the laser emits a light pulse and the test particle is released. At the time the front of the light pulse covers the arc between A-C (arc which is given as L/r0, being L the width of the elevator), I want the difference of r1 vs the radial coordinate of the light pulse, and the difference of r0 vs the radial coordinate of the test particle. For me, that is the distance radially travelled for each of them.

This awful lot of work is to test where the "uniform gravitational field" concept is applicable, if you can do the calculations up to n digits of precision.

Escenario.png
 
  • Like
Likes Dale
  • #5
Ibix said:
You can sanity check your results with Newtonian gravity. Light travels ##l=2\mathrm{m}## in ##\delta t=l/c## seconds. It will rise by approximately ##l^2/2R_E## compared to its initial altitude. The mass will fall approximately ##\frac 12g\delta t^2## in that time. You won't be far wrong with this - how much does it disagree with your detailed calculations?
Sure enough...

For the test particle, the initial guess for the findroot() function is just the Newtonian approximation.

d=pow(tend,2)*G*M/(2*pow(r0,2)) # initial guess
[rend_m] = findroot([lambda r: quad(fmp_dt_dr,[r0-1e-51,r])-tend],(r0-d))

Newton:
0.0000000000000002185311982482473936267804473389293545547985151184677544202782604335814083432271557989293021512927874

Schwarzschild:
0.00000000000000021853119794399465414896083431526218218450276234588121518223277040941841501688951624519599357665033
 
  • Like
Likes Dale
  • #6
PeterDonis said:
...with both staying at the same "vertical" coordinate.
Don't know what this "vertical coordinate" is ¿?. I'm working with the Schwarzschild coordinates. The light pulse, moving from side to side will not maintain a constant radial coordinate, no matter what the "equivalence principle" says. Not even a mathematical straight line will do so, ...is what happens with polar or spherical coordinate systems.

PeterDonis said:
so the "fall distance" relative to the elevator is just the distance the elevator moves upward in the 6.6 nanonseconds...
I don't care the "fall distance" relative to the elevator, that's why the thread is titled "Einstein brother-in-law elevator"... the only think related to an elevator is the width scale of such a thing. Will not enter in discussions about rigid body movement in a non-uniform Gravitational field and so on.

And, of course, if the "Equivalence principle" applies in this case, the spacetime is flat, and why use the Schwarzschild metric? But I do use that metric, and I do obtain those results, being pretty careful as to maintain the required precision solving the equations of motion (that can be wrong, because are handcrafted).

If the difference is in the order of 10-16 m, of course that the number will show as zero with a casio calculator.
 
  • #7
Lluis Olle said:
But I do use that metric,
If you need to use the Schwarzschild metric the tidal effects must be significant, and then we don’t expect the drop within the elevator to be the same for the falling object and the horizontal beam of light.

I’m really at a loss to understand what your point is.
 
  • #8
Nugatory said:
If you need to use the Schwarzschild metric the tidal effects must be significant,
I think he's trying to quantify the tidal effects in a small room on Earth. An easier way is just to hang two plumb lines and measure their failure to be parallel.

The general approach, I think, is to transform the metric to make all first derivatives vanish at the middle of the lift, and then look at the second derivatives times the length scale of the lift.
 
  • #9
Nugatory said:
If you need to use the Schwarzschild metric the tidal effects must be significant, and then we don’t expect the drop within the elevator to be the same for the falling object and the horizontal beam of light.

I’m really at a loss to understand what your point is.
The point is: none.

I just set up an scenario, derive the required Geodesic equations using the Schwarzschild metric, stating at the beginning of the thread that "...the test will be conducted in Earth, simplifying and considering the Schwarzschild metric solution to the Einstein Field Equations, so Earth and surrounding spacetime are not flat."

The Geodesic equations of motion are what they are! Did I got a wrong reasoning in some step, o did some algebraic error, which makes the equations wrong? But if are not wrong, then the final result is what it is, and for the extreme scenario of a 2 meter distance, the light beam will "travel radially" a distance that is 3 times the distance for the test particle. The fact that the scale is 10-16 is ... irrelevant.

What you call "tidal effect", is just the application of the metric, is that easy to understand. That makes the Gravity non uniform, of course.
 
  • #10
Ibix said:
I think he's trying to quantify the tidal effects in a small room on Earth. An easier way is just to hang two plumb lines and measure their failure to be parallel.
That will not give the correct numbers, because in the scenario the test particle do fall radially, but the light beam does not, and is "launched" tangentially.
 
  • #11
Lluis Olle said:
radial coordinate
Why would you even use radial coordinates in this problem? If your "radial coordinate" is measured from the center of the Earth, you are not testing anything related to what you say you're testing:

Lluis Olle said:
This awful lot of work is to test where the "uniform gravitational field" concept is applicable
The Earth's gravitational field is already known not to be "uniform", so it's pointless to use radial coordinates centered on the Earth. You should be using Cartesian coordinates centered on the elevator itself.

Lluis Olle said:
I'm working with the Schwarzschild coordinates.
Then you are not doing what you say you want to do; the "gravitational field" in Schwarzschild coordinates is not "uniform", and your entire analysis is pointless.
 
  • #12
Lluis Olle said:
The point is: none.
In which case, given the fatal flaws I've already pointed out, you are just wasting everyone's time, so this thread is now closed.
 

Similar threads

  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
2
Replies
43
Views
2K
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
8
Views
2K
  • Special and General Relativity
Replies
7
Views
647
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
2
Replies
42
Views
3K
  • Special and General Relativity
Replies
4
Views
1K
  • Special and General Relativity
Replies
9
Views
764
Replies
12
Views
2K
Back
Top